Capacity of a capacitor is `48mu F`. When it is charged from `0.1` C to `0.5` C , change in the energy stored is

Options

A. `2500 J`
B. `2.5 xx 10^(-3) J`
C. `2.5 xx 10^(6) J`
D. `2.42 xx 10^(-2) J`

Answer

A

Explanation

Change in energy `DeltaU = (1)/(2) [(q_(1)^(2) - q_(2)^(2))/(C)]`
`= (1)/(2) [ ((0.5)^(2) - (0.1)^(2))/(48 xx 10^(-6))]`
`= (1)/(2)[ (0.25 - 0.01)/(48 xx 10^(-6))]`
`= (1)/(2) [(0.24)/(48 xx 10^(-6))]`
`= (1)/(2) [ (10^(4))/(2)]`
= 2500 J